Now take the limit (\epsilon \to 0^+):
[ \lim_\epsilon \to 0^+ \frac1\epsilon + i\omega = \frac1i\omega + \pi \delta(\omega) \quad \text(in the sense of distributions) ] fourier transform of heaviside step function
(At (t=0), the value is often taken as (1/2) for symmetry in Fourier analysis, but it’s a set of measure zero, so it doesn’t affect the transform in the (L^2) sense.) The Fourier transform (using the unitary, angular frequency convention) is: Now take the limit (\epsilon \to 0^+): [
[ \hatH(\omega) = \int_-\infty^\infty H(t) , e^-i\omega t , dt = \int_0^\infty e^-i\omega t , dt ] dt = \int_0^\infty e^-i\omega t
[ H(t) = \begincases 1, & t > 0 \ \frac12, & t = 0 \ 0, & t < 0 \endcases ]
Here’s a clear, rigorous explanation of the Fourier transform of the Heaviside step function ( H(t) ), suitable for a textbook, lecture notes, or technical blog. 1. Definition of the Heaviside Step Function The Heaviside step function is defined as:
[ \hatH_\epsilon(\omega) = \int_0^\infty e^-\epsilon t e^-i\omega t , dt = \int_0^\infty e^-(\epsilon + i\omega)t , dt = \frac1\epsilon + i\omega ]